Countinuing where TD! left of, we have that,
Where is the common difference.
Thus,
Becomes,
Thus, (1)
Now for the second equation,
Thus, by equation (1)
Thus,
Simply (2) as,
Thus,
Thus, (2)
Now you have the two equations, solve them,
substitute for from (1),
Thus,
Simplify,
Thus,
Now just check which one works.
Q.E.D.

Furthermore, there is not reason to check which one 'works' since there's no physical interpretation or whatever which has to be checked. All we did was solving the system of equations algebraiclly so both of our solutions are fine, i.e. meet the conditions which were given.

Furthermore, there is not reason to check which one 'works' since there's no physical interpretation or whatever which has to be checked. All we did was solving the system of equations algebraiclly so both of our solutions are fine, i.e. meet the conditions which were given.

I erred again, but it is the method not the answer that is important.
You statement about the checking the two solutions; I know the two solutions work for the algebraic equations but not necessarily for the conditions given for the problem thus we must check.

Well the problem was that knowing we were dealing with an arithmetic sequence, "find a1;a2;a3 if: (the two equations)".
Both solutions are fine, there is no reason to discard one since there's nothing left to check! Unless, of course, totalnewbie didn't give the entire problem (but I think he did or at least I assumed so).